# If $G$ is a Finite Group such that $H\le K$ or $K\le H$ for all Subgroups $H,K$ of $G$, then $G$ is Cyclic and of order $p^n$ for some Prime $p$.

Since each subgroup $K$ is contained in some other subgroup $H$, we can list the subgroups of $G$ in ascending order $$\lbrace 1 \rbrace < G_1< G_2 < G_3\cdots< G_{k-1}<G$$ By Lagrange's Theorem, have have $$1\Big\vert |G_1|, |G_1|\Big\vert |G_2|,\ldots, |G_{k-1}|\Big\vert |G|$$

Denote $\frac{|G_i|}{G_{i-1}}=n_{i-1}$, then $|G|=|G_1|\cdot n_1\cdots n_{k-1}$. How can I deduce that each factor is the same prime $p$ and that $G$ is cyclic?

Cauchy's theorem states that if $p$ is a prime dividing $|G|$ then $G$ has a subgroup of order $p$. If two different primes divide $|G|$ then your condition leads to a contradiction, so $|G| = p^n$ for some $n$.
As for why $G$ is cyclic, consider the subgroup generated by an element in $G \setminus G_{k - 1}$.
The conditions imply that given any two cyclic subgroups, one is a subset of the other. This implies that the group must be cyclic because otherwise there would be two cyclic subgroups that intersect trivially. The same reason implies that the order is a prime power: if there were two relatively prime factors of the order that are both larger than $1$, then there would again be two cyclic subgroups that intersect trivially, hence one could not be a subset of the other. Thus no two nontrivial factors can be relatively prime, so the order is a prime power.